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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: /0 on February 18, 2009, 07:06:59 pm

Title: /0's physics phread
Post by: /0 on February 18, 2009, 07:06:59 pm: I don't know if I've already made a physics thread but in any case either I can't find it or I never made one in the first place :crazy2:

Are the following statements true:

1. A collision is elastic iff $v_2-v_1 = u_1 - u_2$ ; i.e. the departing speed equals the negative of the approaching speeds.

2. A collision is inelastic iff $v_2-v_1 \neq u_1 - u_2$

Also, is 'inelastic' reserved only for collisions where the objects stick together completely, or is it simply defined as not an elastic collision?

Thanks :)
Title: Re: /0's physics phread
Post by: appianway on February 18, 2009, 07:15:29 pm: Elastic and inelastic collisions refer to the conservation of kinetic energy, methinks.
Title: Re: /0's physics phread
Post by: methodsboy on February 18, 2009, 07:20:29 pm: - An elastic collision is where the kinetic energy is conserved throughout the collision
- An inelastic collision is one where some kinetic energy is transformed into heat and sound.
Title: Re: /0's physics phread
Post by: /0 on February 18, 2009, 07:53:18 pm: I know that about energy, but the expressions above are derived from both conservation of momentum and conservation of energy. It would be much more efficient if I could use those expressions to determine the elasticity of a collision, rather than calculating the KE before and after, and I want to know if it is acceptable.
Title: Re: /0's physics phread
Post by: appianway on February 18, 2009, 08:27:02 pm: Just keep in mind that momentum's a vector and that energy's a scalar quantity (thus the negative in your first equation shouldn't exist).
Title: Re: /0's physics phread
Post by: /0 on February 22, 2009, 02:35:24 am: Another:

1. A golf ball rolls at 4m/s at the top of a small hill of radius 2m. Will the ball lose contact with the surface? Show your calculations and reasoning. A force diagram is advised.

I have proved that at the top of the hill the ball is still in contact, but could it lose contact on the sides of the hill? The centripetal force there would be smaller, wouldn't it?

2. In vertical circular motion, is the speed still constant? Or is it greater at the bottom than at the top? If so, then does $\frac{mv^2}{r}$ change with time?
What information do you need to find the speeds at different times?

3. What is the action-reaction pair to the centripetal force?

4. What is the advantage of lowering the centre of mass when cornering?
Title: Re: /0's physics phread
Post by: /0 on February 22, 2009, 04:33:14 pm: Another two:

5. Calculate the gravitational force attracting two masses of 1 kg, separated by 0.40m. Take $G = 6.67 \times 10^{-11} m^3kg^{-1}s^2$

I used the formula $\frac{GMm}{r^2}$ but I'm wondering, does that formula only give the force on one body by the other ?

So would the total gravitational force attracting be the force on both bodies = $\frac{2GMm}{r^2}$ ? Thanks

6. A rocket is disabled at a height of 9 Earth radii above the surface of the earth. It is moving very slowly. By how much will its velocity change in the next 200s? (Take g at the surface of the earth as 10m/s, and assume that the local value of g does not change in the 200s)
Title: Re: /0's physics phread
Post by: TrueTears on February 22, 2009, 05:13:05 pm: Quote from: /0 on February 22, 2009, 04:33:14 pm
Another two:

5. Calculate the gravitational force attracting two masses of 1 kg, separated by 0.40m. Take $G = 6.67 \times 10^{-11} m^3kg^{-1}s^2$

I used the formula $\frac{GMm}{r^2}$ but I'm wondering, does that formula only give the force on one body by the other ?

So would the total gravitational force attracting be the force on both bodies = $\frac{2GMm}{r^2}$ ? Thanks

6. A rocket is disabled at a height of 9 Earth radii above the surface of the earth. It is moving very slowly. By how much will its velocity change in the next 200s? (Take g at the surface of the earth as 10m/s, and assume that the local value of g does not change in the 200s)

i'll have a go at 6.

Looking at the formula $g = \frac{GM}{R^2}$ , let GM = k so $g = \frac{k}{R^2}$

$k = gR^2$

so we have g = 10 and R = 1R when the rocket is at the surface this gives k = 10 x $1R^2<br />$
later when the rocket is at 9 Earth radii we have $k = g (10R)^2$ $k = g \times 100R^2$ (notice its 10 because you have to add the earth radius as well)

equate the 2 k's and we have $10R^2 = 100R^2g$ therefore g = 0.1.

So we know the a = 0.1 t = 200 and u = 0. By subbing into the equation v = u + at we get v = 20 $ms^{-1}$
Title: Re: /0's physics phread
Post by: TrueTears on February 22, 2009, 05:16:20 pm: For question 5, the formula only gives the force on one body by the other, but by Newton's third Law, the other body would exert an equal force on the other object. So you don't need to do
$2 \times \frac{GMm}{r^2}$
Title: Re: /0's physics phread
Post by: TrueTears on February 22, 2009, 05:25:27 pm: Quote from: /0 on February 22, 2009, 02:35:24 am

Another:
2. In vertical circular motion, is the speed still constant? Or is it greater at the bottom than at the top? If so, then does $\frac{mv^2}{r}$ change with time?
What information do you need to find the speeds at different times?

For question 2, in vertical circular motion speed is not constant. Effects by gravity causes the object to go slower at the top than at the bottom. $\frac{mv^2}{r}$ does change when the object is at the top compared to when its at the bottom. When the object is at the top, you have its Normal force acting down towards the centre, the force due to gravity acting down and $\frac{mv^2}{r}$ acting down towards the centre. $N + F_g = F_{net}$ . Say you were in a roller coaster, and you were at the top of a circular track, you would feel slightly lighter.

But if the object was on the bottom, Its normal force is acting up towards the centre, the force due to gravity is acting down and the $\frac{mv^2}{r}$ is acting up towards the centre, so you would get $N - F_g = F_{net}$ . Again if you were in a roller coaster, and you were at the bottom of the circular track, you would feel slightly heavier.
Title: Re: /0's physics phread
Post by: TrueTears on February 22, 2009, 05:33:08 pm: For question 3:

All forces have action-reaction pairs. But the NET force on an object is not the same as a force on the object. Centripetal force is the force pulling the object towards the centre of the curve, and this force is the net force on that object, so it doesn't have a opposite force.

lets say 2 people are pulling a box in opposite directions. One with 2 newtons of force and the other with 6 newtons of force. The net force is 4 newtons in the direction of the 6 newton force. There is no "equal and opposite" force to this 4 newton net force.
Title: Re: /0's physics phread
Post by: /0 on February 22, 2009, 05:39:37 pm: thanks truetears
Title: Re: /0's physics phread
Post by: TrueTears on February 22, 2009, 05:56:13 pm: Also i think lowering the centre of mass has something to do with gravity...? Since gravity acts from the centre of mass, and lowering the centre of mass while turning a corner would 'pin' you more to the ground? lol i dono someone clear this up for me.
Title: Re: /0's physics phread
Post by: /0 on March 24, 2009, 11:31:55 am: hmmm

If a car is moving at 40m/s and a truck collides with it, stopping the car, would you say the work the truck does on the car is negative or positive? The answers say it's positive but how could that be, seeing as the car's change in kinetic energy is negative?

When you say work is done ON an object what do you mean?
e.g. If a truck hits a car, stopping it, is it the work exerted by the truck, or the work done [/i]to the system which is the car?
Title: Re: /0's physics phread
Post by: pHysiX on March 24, 2009, 06:01:35 pm: hmmm...here's my theory:

the way i'm thinking, if you take the reference from the car then yes, the work done by the truck is negative. however, work is an energy, hence, getting negative energy is "invalid". the direction really does not matter in this case because the work done is how far the truck has moved and the force it has applied. if you think about it, it's actually a negative distance and a negative net force by the truck, hence positive.

well that's my random thoughts...not sure how right or how wrong i am. open to more suggestions =]
Title: Re: /0's physics phread
Post by: Mao on March 24, 2009, 08:17:44 pm: The force on the car is opposite the displacement of the car, the work done by the truck on the car is negative (that is, the car did work on the truck).
Title: Re: /0's physics phread
Post by: /0 on May 30, 2009, 05:42:25 pm: Hey does anyone know what the units for $\tau$ are? Are they meant to be $Nm$ or $mN$ ?
Title: Re: /0's physics phread
Post by: evaporade on May 30, 2009, 08:51:09 pm: Quote from: /0 on March 24, 2009, 11:31:55 am
hmmm

If a car is moving at 40m/s and a truck collides with it, stopping the car, would you say the work the truck does on the car is negative or positive? The answers say it's positive but how could that be, seeing as the car's change in kinetic energy is negative?

When you say work is done ON an object what do you mean?
e.g. If a truck hits a car, stopping it, is it the work exerted by the truck, or the work done [/i]to the system which is the car?

From the moment the truck comes in contact with the car until it stops, the truck moves forwards and the force exerted by the truck on the car is also in the forward direction. So the work done by the truck is positive because the displacement of the truck (and the car) and the force of the truck are in the same direction.
Title: Re: /0's physics phread
Post by: evaporade on May 31, 2009, 10:30:12 am: In fact the car and the truck question is quite complicated. The answer depends on the mass of the car, mass of the truck, velocity of the truck and whether friction is involved.

If there is no friction involved, the car stops when the momentum of the truck is exactly equal and opposite the momentum of the car. In this scenario, the work done by the truck is negative in terms of reducing the kinetic energy of the car because the force of the truck and the displacement of the car (centre of mass) are opposite in direction.

In a real situation the momentum of the truck is greater than the opposite momentum of the car and friction is involved. The answer in my last post applies.
Title: Re: /0's physics phread
Post by: kamil9876 on May 31, 2009, 01:18:29 pm: Evaporade: I think your definition of work is correct(I checked on wikipedia), however you are only considering half the story.

When the car is struck by the truck the work the truck does on the car is negative, however once it turns around the work is positive.
This situation can be compared to a ball being thrown up in the air. Initially grativy does negative work because the displacement is up while the force is acting down, however after the turnaround the work is positive since the direction the ball is falling is down and the force is also down.
In the truck case the car's direction is still positive for som time $\delta t$ however beyond that it begins to turn around (just like the ball does) and so the direction of car and force are both negative hence positive work.
This is consistent with /0's claim concerning the change in kinetic energy: car slows down, then speeds up after turn around(just like ball does)

Key: $W=F.d=|F||d|cos\theta$ where d is displacement of object work is being done on while F is the force being applied.

Edit: my claim about the car's direction being positive for some small time can be justified as follows: the velocity of the car is +40m/s, however it eventually becomes negative but the change in velocity must be continous because if it wasn't then we would need infinite force.(assuming mass of car is constant: $F_{Average}=m\frac{\Delta V}{\Delta t}$ So a change of over 40m/s in v in an instant needs unrealistically large force)
Title: Re: /0's physics phread
Post by: methodsboy on May 31, 2009, 01:58:33 pm: Quote from: /0 on May 30, 2009, 05:42:25 pm
Hey does anyone know what the units for $\tau$ are? Are they meant to be $Nm$ or $mN$ ?
N/m
Title: Re: /0's physics phread
Post by: evaporade on May 31, 2009, 03:14:32 pm: I considered the whole story that is why I said 'The answer depends on the mass of the car, mass of the truck, velocity of the truck and whether friction is involved.' The 'overall' work (+ or -) done by the truck on the car depends on its momentum before collision. 'Overall' work was discussed because it is what years 11 and 12 required to know, not moment to moment analysis of the motion.
Title: Re: /0's physics phread
Post by: evaporade on May 31, 2009, 03:26:23 pm: 'my claim about the car's direction being positive for some small time can be justified as follows: the velocity of the car is +40m/s, however it eventually becomes negative but the change in velocity must be continous because if it wasn't then we would need infinite force.(assuming mass of car is constant: F_{Average}=m\frac{\Delta V}{\Delta t} So a change of over 40m/s in v in an instant needs unrealistically large force)'

I think you are talking nonsense here. For sure the centre of mass of the car slows down gradually due to the collapse of the front end of the car. You don't need 'unrealistically large force' to stop the car.
Title: Re: /0's physics phread
Post by: evaporade on May 31, 2009, 03:40:45 pm: I better stop here, otherwise I may end up like the person in /0 's picture. This is to be expected on this forum.
Title: Re: /0's physics phread
Post by: kamil9876 on May 31, 2009, 05:15:50 pm: Quote from: evaporade on May 31, 2009, 03:26:23 pm

I think you are talking nonsense here. For sure the centre of mass of the car slows down gradually due to the collapse of the front end of the car. You don't need 'unrealistically large force' to stop the car.

I didn't say that, of course the car can be stopped, and i said that it is being stopped gradually(my use of the word 'continously') i just said that being stopped in an 'instant' is impossible meaning that it cannot go from 40m/s to some negative velocity without going through all the velocities in between. Now because of that, it is still going forward for some time $\delta t$ (and so negative work in this duration).

Btw, i don't mind you saying that what i said was nonsense so I give you the right to continue criticizing any of my nonsense.
Title: Re: /0's physics phread
Post by: /0 on May 31, 2009, 06:04:17 pm: Oh dear, I think this is getting a bit beyond VCE
So you're saying, if the car starts the collision moving in one direction, but the dollision makes the car move in the other drection, then you have both positive and negative work on the car? How would you deal with that?
Title: Re: /0's physics phread
Post by: kamil9876 on May 31, 2009, 07:48:30 pm: well its NOT positive AND negative at the same time. Think about the ball being thrown in the air scenario. On the path up its negative, and the path down it's positive. Negative for path up because its a dot product of two vectors opposite to each other. Otherwise for path down.

In the car situation, the moment of time the car is being stopped, it is slowing down and so it's analogous to the path up situation (force is acting opposite to the initial tiny displacement of the car). But once the car begins moving backwards, well that is analogous to the path down situation.
Title: Re: /0's physics phread
Post by: kamil9876 on May 31, 2009, 07:57:20 pm: and yes, i think this is beyond yr12 since it is formalised with the dot product(and math in yr12 physics is nothing beyond manipulating linear equations). THe 'negative work because kinetic energy decreased' argument get's you the same answer anyway.

I think the answer to this question depends on whether you are considering the work it takes just to stop the car or if the car is being dragged back by the truck(they squashed into one wreck) and you want the work done when the car is being dragged back by the truck.
Title: Re: /0's physics phread
Post by: /0 on May 31, 2009, 08:41:35 pm: Thanks kamil and evaporade !

For these electrical devices, which must use AC, which must use DC and which can use either:
Resistor
Thermistor
LED
LDR
Photodiode
Phototransistor
Voltage Amplifier

thx
Title: Re: /0's physics phread
Post by: /0 on June 07, 2009, 01:27:19 am: I am having lots of problems with the Insight 2009 Trial Exam...
http://vcenotes.com/forum/index.php/topic,8214.15.html

Question 6 - Electronics and Photonics:
If you look at the transfer characteristic of the amplifier, a voltage of $1.5mV$ should give an output voltage of something like $8.2V$ , and a voltage of $-1.5mV$ should give $10V$ , because that is well within the saturation region of the amplifier.
But in the answer they just apply the linear gain to the max and min of the input graph... How is this correct if it defies the characteristic graph??
I would expect an output graph that is mostly constant at 10V but which drops down every now and then.

Question 11 - Electronics and Photonics:
Are the answers wrong with this!?? Should the modulated signal be modulated at its minimums as well? They just have it at constant minimum!

Question 5 - Structures and Materials:
Isn't D as acceptable an answer as A?
Title: Re: /0's physics phread
Post by: TrueTears on June 07, 2009, 01:31:12 am: Quote from: /0 on June 07, 2009, 01:27:19 am
I am having lots of problems with the Insight 2009 Trial Exam...
http://vcenotes.com/forum/index.php/topic,8214.15.html

Question 6 - Electronics and Photonics:
If you look at the transfer characteristic of the amplifier, a voltage of $1.5mV$ should give an output voltage of something like $8.2V$ , and a voltage of $-1.5mV$ should give $10V$ , because that is well within the saturation region of the amplifier.
But in the answer they just apply the linear gain to the max and min of the input graph... How is this correct if it defies the characteristic graph??
I would expect an output graph that is mostly constant at 10V but which drops down every now and then.

Question 11 - Electronics and Photonics:
Are the answers wrong with this!?? Should the modulated signal be modulated at its minimums as well? They just have it at constant minimum!

Question 5 - Structures and Materials:
Isn't D as acceptable an answer as A?

I remember Q 6 and thought exactly as you did, but if you look closely the graph for $V_{out}$ they give you begins at 0. A suitable graph would have started at around 5 as the mid point for the $V_{out}$ graph and then oscillate to 10 and down to 1.4. So basically you take 0 on the $V_{out}$ graph as 5, and the highest you go up is around 4 which corresponds to around 10 if you had the midpoint around 5.
Title: Re: /0's physics phread
Post by: /0 on June 07, 2009, 01:57:09 am: Quote from: TrueTears on June 07, 2009, 01:31:12 am
Quote from: /0 on June 07, 2009, 01:27:19 am
I am having lots of problems with the Insight 2009 Trial Exam...
http://vcenotes.com/forum/index.php/topic,8214.15.html

Question 6 - Electronics and Photonics:
If you look at the transfer characteristic of the amplifier, a voltage of $1.5mV$ should give an output voltage of something like $8.2V$ , and a voltage of $-1.5mV$ should give $10V$ , because that is well within the saturation region of the amplifier.
But in the answer they just apply the linear gain to the max and min of the input graph... How is this correct if it defies the characteristic graph??
I would expect an output graph that is mostly constant at 10V but which drops down every now and then.

Question 11 - Electronics and Photonics:
Are the answers wrong with this!?? Should the modulated signal be modulated at its minimums as well? They just have it at constant minimum!

Question 5 - Structures and Materials:
Isn't D as acceptable an answer as A?

I remember Q 6 and thought exactly as you did, but if you look closely the graph for $V_{out}$ they give you begins at 0. A suitable graph would have started at around 5 as the mid point for the $V_{out}$ graph and then oscillate to 10 and down to 1.4. So basically you take 0 on the $V_{out}$ graph as 5, and the highest you go up is around 4 which corresponds to around 10 if you had the midpoint around 5.

Why would $V_{out}$ start at 5? Shouldn't it start at 10? If you graph it like that doesn't that defy the characteristic graph?
Title: Re: /0's physics phread
Post by: TrueTears on June 07, 2009, 01:59:04 am: Quote from: /0 on June 07, 2009, 01:57:09 am
Quote from: TrueTears on June 07, 2009, 01:31:12 am
Quote from: /0 on June 07, 2009, 01:27:19 am
I am having lots of problems with the Insight 2009 Trial Exam...
http://vcenotes.com/forum/index.php/topic,8214.15.html

Question 6 - Electronics and Photonics:
If you look at the transfer characteristic of the amplifier, a voltage of $1.5mV$ should give an output voltage of something like $8.2V$ , and a voltage of $-1.5mV$ should give $10V$ , because that is well within the saturation region of the amplifier.
But in the answer they just apply the linear gain to the max and min of the input graph... How is this correct if it defies the characteristic graph??
I would expect an output graph that is mostly constant at 10V but which drops down every now and then.

Question 11 - Electronics and Photonics:
Are the answers wrong with this!?? Should the modulated signal be modulated at its minimums as well? They just have it at constant minimum!

Question 5 - Structures and Materials:
Isn't D as acceptable an answer as A?

I remember Q 6 and thought exactly as you did, but if you look closely the graph for $V_{out}$ they give you begins at 0. A suitable graph would have started at around 5 as the mid point for the $V_{out}$ graph and then oscillate to 10 and down to 1.4. So basically you take 0 on the $V_{out}$ graph as 5, and the highest you go up is around 4 which corresponds to around 10 if you had the midpoint around 5.

Why would $V_{out}$ start at 5? Shouldn't it start at 10? If you graph it like that doesn't that defy the characteristic graph?
No, I'm saying if you had to draw a graph without the axes and scales given in the exam. If you had to draw on yourself you'd have mid point of the sin curve as y = 5V, then the highest would go up to around 10V ish and lowest point be would be around 1.4V.

Now just consider everything being moved down 5, your mid point would be lowered to 0, your highest value would be around 4-5ish (depending what value you took before) and your lowest point(1.4V) would be lowered to around -4ish. Now it fits the scales and axes given to you.
Title: Re: /0's physics phread
Post by: /0 on June 07, 2009, 04:27:42 am: lol true, are we allowed to do that though?
Title: Re: /0's physics phread
Post by: TrueTears on June 07, 2009, 12:57:03 pm: Quote from: /0 on June 07, 2009, 04:27:42 am
lol true, are we allowed to do that though?
Yeah, just transforming the whole graph down by 5 units, the amplitude and frequency stays the same.

It's the same for characteristics graphs, where the midpoint doesn't always have to start at (0,0).
Title: Re: /0's physics phread
Post by: /0 on June 07, 2009, 01:58:37 pm: thxxx :P
Title: Re: /0's physics phread
Post by: dejan91 on June 07, 2009, 02:28:13 pm: But in this case, it doesn't apply because they've already given you a scale. Which is stupid because their answer doesn't fit with the graph or scale provided....
Title: Re: /0's physics phread
Post by: /0 on June 08, 2009, 06:07:59 pm: For this LED, what is the threshold voltage?

(http://img246.imageshack.us/img246/6524/ledanu.jpg)
Title: Re: /0's physics phread
Post by: TrueTears on June 08, 2009, 06:10:23 pm: Correct me if I'm wrong, but I don't think you can determine it. They'd have to give you a certain current and you'd have to read the voltage off the graph.
Title: Re: /0's physics phread
Post by: /0 on June 08, 2009, 06:11:41 pm: If you wanna check this question up it's from question 7 in the electronics section of Physics TSFX 2009

http://vcenotes.com/forum/index.php/topic,8214.30.html
Title: Re: /0's physics phread
Post by: dejan91 on June 08, 2009, 08:10:27 pm: I didn't get that one either. Realistically, it could be anythnig from 1.5-2.but in the answers, it says anything from 1.8-1.9 is acceptable. Yeah right...
Title: Re: /0's physics phread
Post by: /0 on June 18, 2009, 10:11:30 pm: Is it possible to express a magnetic field as a function like $B(x,y)$ for a paritcular magnet strength... or perhaps if it's not a function then can the strength of the field be found at any particular point mathematically in general?Nevermind: $B = \frac{\mu_0 I}{2\pi R}$

Also, say... if you were a north pole and you were momentarily at ~~on the south pole~~ a distance $d \neq 0$ away from the north pole of another magnet. Is there an 'escape velocity' at which you can escape the attraction of the other magnet, or would it require infinite energy to do that?
Title: Re: /0's physics phread
Post by: /0 on June 19, 2009, 07:58:53 pm: How does AC current actually work? If the current goes back and forth then how does it get anywhere?
Title: Re: /0's physics phread
Post by: /0 on June 19, 2009, 10:15:36 pm: Calculate the maximum magnetic flux passing through a coil of 50 turns, 12cm^2 in area in a magnetic field of strength 0.025T.

I just got $(0.025)(12 \times 10^{-4}) = 3 \times 10^{-5} Tm^2$ . Shouldn't flux be independent of the number of coils? Doesn't it just depend on area?
Title: Re: /0's physics phread
Post by: TrueTears on June 19, 2009, 10:20:42 pm: No, you must consider how many turns there are.

Consider a solenoid, if you just have 1 "turn" of coil, let the area be A $m^2$

But if you have more "turns" the coil gets bigger (or rather longer) right? So you are exposed to more area. Say you got $x$ number of turns, your area would now be $x \times A$ $m^2$
Title: Re: /0's physics phread
Post by: /0 on June 19, 2009, 10:35:18 pm: thanks tt

also,

"A magnet falling through a metal tube can achieve terminal velocity. Why?"

Is this an apt description?

The magnet induces a magnetic field in the metal of the tube. Due to the magnet's velocity relative to the external magnetic field, its charges are separated. The seaparation of charges also means that, by the right-hand slap rule, the external magnetic field exerts an upward force on the magnet. When the speed of the magnet reaches a certain point, the upwards magnetic forces balances the gravitational force, so terminal velocity is achieved.

Is there a better way to explain it?
Title: Re: /0's physics phread
Post by: TrueTears on June 19, 2009, 10:38:24 pm: Yeap, that's fine, just remember that also as the rod falls faster and faster the magnetic force on it increases (F = Bqv). The rod will achieve terminal velocity once the magnetic force equals the weight force.

Other than that, your explanation is fine. Very detailed.
Title: Re: /0's physics phread
Post by: /0 on June 19, 2009, 10:44:42 pm: thanks again TT :D

If you have a circular loop completely immersed in a magnetic field $\otimes$ and you move it out of the field.
Inside the loop, $\Delta \Phi = \odot$
But outside the loop, $\Delta \Phi = \odot$ as well...

Why are we only concerned with the $\Delta \Phi$ inside the loop?

NVM WORKED IT OUT
Title: Re: /0's physics phread
Post by: TrueTears on June 19, 2009, 10:48:40 pm: Quote from: /0 on June 19, 2009, 10:44:42 pm
thanks again TT :D

If you have a circular loop completely immersed in a magnetic field $\otimes$ and you move it out of the field.
Inside the loop, $\Delta \Phi = \odot$
But outside the loop, $\Delta \Phi = \odot$ as well...

Why are we only concerned with the $\Delta \Phi$ inside the loop?
What do you mean by loop? loop of wire? loop with current through it?
Title: Re: /0's physics phread
Post by: Mao on June 20, 2009, 12:14:10 am: Quote from: /0 on June 19, 2009, 07:58:53 pm
How does AC current actually work? If the current goes back and forth then how does it get anywhere?

Electric power comes from motion of electrons. In this case, the overall motion of electrons in AC current is nought, but it can be described as oscillations, which is still a type of motion, hence there is electric power.
Title: Re: /0's physics phread
Post by: /0 on June 20, 2009, 12:22:07 am: Quote from: Mao on June 20, 2009, 12:14:10 am
Quote from: /0 on June 19, 2009, 07:58:53 pm
How does AC current actually work? If the current goes back and forth then how does it get anywhere?

Electric power comes from motion of electrons. In this case, the overall motion of electrons in AC current is nought, but it can be described as oscillations, which is still a type of motion, hence there is electric power.

woah that's sick as, thx man

Quote from: TrueTears on June 19, 2009, 10:48:40 pm
Quote from: /0 on June 19, 2009, 10:44:42 pm
thanks again TT :D

If you have a circular loop completely immersed in a magnetic field $\otimes$ and you move it out of the field.
Inside the loop, $\Delta \Phi = \odot$
But outside the loop, $\Delta \Phi = \odot$ as well...

Why are we only concerned with the $\Delta \Phi$ inside the loop?
What do you mean by loop? loop of wire? loop with current through it?

Just like Q8 Page 274

I was wrong... outside the loop $\Delta \Phi = \otimes$

Another:

Two coils are placed one on top of the other with their centres in line.
(a) If a battery is switched on in the bottom coil, producing a clockwise current as seen from above, what happens in the top coil?
(b) Would the effect be different if the battery was connected to the top coil?
(c) Would the effect be different if the battery was switched off?

Are the answer (a) nothing (b) no (c) no ?
Because in (a) the top coil isn't moving so i would think that means there's no force?
And in (b) The bottom coil isn't moving so the top coil does no force on the bottom?
And in (c) nothing happens anyway?
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 12:23:26 am: Oh okay, here's how I did the question.

$\Delta \phi = \phi_{final} - \phi_{initial} =$ negative number (Since $\phi_{initial} > \phi_{final}$ )

Let into page be +ve.

So $\Delta \phi$ must be out of page. Induced magnetic field must be into the page (opposite direction)

Using right hand grip rule, current goes clockwise :)
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 12:56:16 am: Quote from: /0 on June 20, 2009, 12:22:07 am
Quote from: Mao on June 20, 2009, 12:14:10 am
Quote from: /0 on June 19, 2009, 07:58:53 pm
How does AC current actually work? If the current goes back and forth then how does it get anywhere?

Electric power comes from motion of electrons. In this case, the overall motion of electrons in AC current is nought, but it can be described as oscillations, which is still a type of motion, hence there is electric power.

woah that's sick as, thx man

Quote from: TrueTears on June 19, 2009, 10:48:40 pm
Quote from: /0 on June 19, 2009, 10:44:42 pm
thanks again TT :D

If you have a circular loop completely immersed in a magnetic field $\otimes$ and you move it out of the field.
Inside the loop, $\Delta \Phi = \odot$
But outside the loop, $\Delta \Phi = \odot$ as well...

Why are we only concerned with the $\Delta \Phi$ inside the loop?
What do you mean by loop? loop of wire? loop with current through it?

Just like Q8 Page 274

I was wrong... outside the loop $\Delta \Phi = \otimes$

Another:

Two coils are placed one on top of the other with their centres in line.
(a) If a battery is switched on in the bottom coil, producing a clockwise current as seen from above, what happens in the top coil?
(b) Would the effect be different if the battery was connected to the top coil?
(c) Would the effect be different if the battery was switched off?

Are the answer (a) nothing (b) no (c) no ?
Because in (a) the top coil isn't moving so i would think that means there's no force?
And in (b) The bottom coil isn't moving so the top coil does no force on the bottom?
And in (c) nothing happens anyway?
Okay here is how I did the question.

a) (http://img23.imageshack.us/img23/3236/q9jarcaranda.jpg)

Bird's eye view of the diagram.

Now, lower coil the magnetic field is into the page, using RHGR

now there must be a induced magnetic field in the upper coil. First lets find the direction of change in flux.

let into the page be +ve

lets work out change in flux for lower coil

flux final - flux initial, it is positive since flux final > flux initial, so that means induce mag field is opposite direction, so its out of page, now if it is out of page lets use RHGR on upper coil

Therefore current is anti clockwise for upper coil.
Title: Re: /0's physics phread
Post by: /0 on June 20, 2009, 02:19:27 am: OH, so we are looking at the short time between the battery being off and the battery being on, cool thanks bro

But the current in the top wire should only be for a split second, right? Because as soon as the battery is on the magnetic field will become constant?
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 02:24:28 am: Yeap that's right :)
Title: Re: /0's physics phread
Post by: /0 on June 20, 2009, 11:25:07 pm: thxthxthxthxthxthxthxthxthxthxTT

How can a motor operate as a DC generator?
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 11:25:54 pm: I thought generators are AC lol
Title: Re: /0's physics phread
Post by: /0 on June 20, 2009, 11:27:04 pm: Hmmm that's interesting. it's Q15 pg 274
Might have to ask the BMW
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 11:29:28 pm: Quote from: /0 on June 20, 2009, 11:27:04 pm
Hmmm that's interesting. it's Q15 pg 274
Might have to ask the BMW
Hmmm I think maybe change the split rings to slip rings? Dono but that makes a AC generator lols.
Title: Re: /0's physics phread
Post by: /0 on June 20, 2009, 11:51:37 pm: OK, I have an interesting conundrum...

Say you have a rectangular wire loop lying horizontal which is COMPLETELY immersed in an infinite, uniform, perpendicular magnetic field.

If you move the wire, according to F = Bqv, the charges should separate, and hence there must be an emf.

However, according to Lenz's Law, no matter where you move this wire, the flux through the wire will always be the same, because the magnetic field is uniform. Thus, there should be no emf.

How can this be resolved?
Title: Re: /0's physics phread
Post by: TrueTears on June 20, 2009, 11:56:30 pm: By moving the wire, you are only changing the DIRECTION of the electrons not the speed, v stays the same.
Title: Re: /0's physics phread
Post by: /0 on June 21, 2009, 12:06:10 am: Quote from: TrueTears on June 20, 2009, 11:56:30 pm
By moving the wire, you are only changing the DIRECTION of the electrons not the speed, v stays the same.

Hmmm? Wouldn't v change?

Ok let's say the wire is already in motion, moving at a constant speed v perpendicular to the field. Is there an emf then?
Title: Re: /0's physics phread
Post by: TrueTears on June 21, 2009, 12:07:10 am: Is the wire moving in a straight line or rotating?

Say the wire is moving up, with infinite long uniform magnetic field around it (say left of it is north and right of it is south), yes there would be a force on the wire, this means the electrons are moving up, using the left hand slap rule we get the forcing facing you (out of the page), as in the wire is being pushing towards you. There is no emf because the electrons are not separated on either end of the wire.

If it was rotating then yes there would be a induced emf due to change of flux.
Title: Re: /0's physics phread
Post by: /0 on June 21, 2009, 12:08:34 am: Just moving in a straight line
Title: Re: /0's physics phread
Post by: TrueTears on June 21, 2009, 12:11:10 am: Edited last post, check.
Title: Re: /0's physics phread
Post by: /0 on June 21, 2009, 12:19:12 am: Quote from: TrueTears on June 21, 2009, 12:07:10 am
Is the wire moving in a straight line or rotating?

Say the wire is moving up, with infinite long uniform magnetic field around it (say left of it is north and right of it is south), yes there would be a force on the wire, this means the electrons are moving up, using the left hand slap rule we get the forcing facing you (out of the page), as in the wire is being pushing towards you. There is no emf because the electrons are not separated on either end of the wire.

If it was rotating then yes there would be a induced emf due to change of flux.

Wait, sorry what I mean is if the magnet is as it is in question 11 Page 274, but imagine the dots go on till infinity.

Now if the wire moves in the direction of A, is there an emf?
Title: Re: /0's physics phread
Post by: TrueTears on June 21, 2009, 12:20:25 am: In response to your post before you edited/deleted it.

If the wire was coming out of the page that would be impossible wouldn't it?

You initial condition was that the wire was surrounded by a magnetic field, it can only move up (considering the magnetic field was like a hallow tube, and the wire fits in the hallow tube)

If it comes out of the page it would just hit the side of the hallow tube?

If only the "sides" of the wire was surrounded by magnetic fields and the wire comes out of the page, then yes there would be an emf, check with Lenz Law, it also proves there to be a induced emf :)
Title: Re: /0's physics phread
Post by: /0 on June 21, 2009, 12:25:38 am: Sorry I was edited my previous post to show what I meant.

Also in response to your previous post, if the poles of the magnet were at a very long distance away, I don't see why the wire wouldn't be able to move in any direction it liked.
Title: Re: /0's physics phread
Post by: TrueTears on June 21, 2009, 12:28:46 am: If the poles of the magnets were very far away, the magnetic field at the wire would be so weak that it'd approach 0?
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 12:22:32 am: lol thx tt but don't worry about that it's a stupid question

Another though, how does decreasing the area of the rotating coil decrease the speed of the motor?
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 12:29:04 am: Reducing the area reduces 2 things, the distance to the axle of rotation and the length of the coil.

the rotational torque is dependent on the length to the axle of rotation and force applied.

smaller length of coil implies smaller force which implies smaller torque which implies slower

smaller distance to axle of rotation implies smaller torque which implies slower.
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 12:38:01 am: pr0 thx man
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 12:44:06 am: A current is flowing in the coil between the magnets the coil is rotating like a motor side AB of the coil is rotating towards us.

Which of the following best describes the current in the coil?
A. It is flowing from A to B along this side.
B. It must be alternating.
C. It is flowing from B to A along this side.
D. It could be flowing in either direction, one cannot tell.

:) (: :) (:
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 12:26:42 pm: I thought answer would be just C.

I just used the right hand slap rule.

the force on BA is coming out of the page, magnetic field is from right to left. Your fingers point up, hence current flows from B to A.

Why must it be alternating? I thought the diagram could represent DC motor or AC/DC generator [since it doesn't show slip rings or split ring commutator]. But DC motor uses DC current and DC generator is pulsating DC current, so it may not be alternating? (even though the commutator reverses the current in a DC motor, the +ve and -ve terminals don't switch.)
Title: Re: /0's physics phread
Post by: Mao on June 24, 2009, 02:06:11 pm: Quote from: TrueTears on June 24, 2009, 12:26:42 pm
I thought answer would be just C.

I just used the right hand slap rule.

the force on BA is coming out of the page, magnetic field is from right to left. Your fingers point up, hence current flows from B to A.[/tex]

my apologies, I thought it was a generator

Quote from: TrueTears on June 24, 2009, 12:26:42 pm
Why must it be alternating? I thought the diagram could represent DC motor or AC/DC generator [since it doesn't show slip rings or split ring commutator]. But DC motor uses DC current and DC generator is pulsating DC current, so it may not be alternating? (even though the commutator reverses the current in a DC motor, the +ve and -ve terminals don't switch.)

The current in the coil must always be alternating (every half turn) to maintain the torque in the same direction. This is the purpose of a split-ring commutator, to alternate the current inside the coil.
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 02:08:04 pm: Quote from: TrueTears on June 24, 2009, 12:26:42 pm
Why must it be alternating? I thought the diagram could represent DC motor or AC/DC generator [since it doesn't show slip rings or split ring commutator]. But DC motor uses DC current and DC generator is pulsating DC current, so it may not be alternating? (even though the commutator reverses the current in a DC motor, the +ve and -ve terminals don't switch.)

The current in the coil must always be alternating (every half turn) to maintain the torque in the same direction. This is the purpose of a split-ring commutator, to alternate the current inside the coil.

Ahh yeap, thanks that's what I meant :)
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 07:48:21 pm: thx

also in VCAA 2003, AOS 2 Question 10 http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/physics12003.pdf or Checkpoints Q108

Why isn't C a possible option alongside B?
Designating N to S as positive shouldn't really make a difference should it? After all, isn't that designation independent of the wire's position? I thought starting with the flux as positive is pretty arbitrary.
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 07:58:41 pm: Ahhh I remember this tricky fellow :P

Now, we know the current-time graph has same frequency and looks the same as the emf-time graph cept just with different amplitude.

Now, the graph is a bit tricky in checkpoints but the exam paper makes it much clearer.

The "bottom" wire is connected to the last slip ring and the "top" wire is connected to the first slip ring. (the one "closer" to X)

Now lets use Lenz Law on the coil.

As the coil rotates clockwise, flux gets smaller, so change in flux is negative hence its from S to N direction, so induced mag field must be from N to S, use RHG rule and we see current goes from X to W to V to U.

BUT! the question wants current from U to V. So normally you would have A as your answer right [because that represents V to U NOT U to V]? But this time we want the - (-derivative) of the flux-time graph. Hence we just need the "derivative" graph, which is clearly D.
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 08:00:52 pm: err thansk but that's 109
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 08:01:53 pm: Oh shit, soz I thought you said Q 109 LOL.

Anyways for Q 108, C is clearly wrong because it says N to S is positive.

Yes direction for change in flux is arbitrary, just plain flux direction is also arbitrary, but clearly if it says N to S is positive it means the first "face" that the mag field hits yields a POSITIVE flux reading.

[Hard to explain just by typing but imagine the square coil with 2 "faces" the diagram depicts the mag field first hitting the "left face" and then passing through to the "right face", clearly if N to S is positive, the flux would be positive for the "left face" because that what the mag field hits first and as a result the "right face" is negative]

C would depict the flux-time graph if N to S was NEGATIVE. Ie, hitting the "right face" first.
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 08:04:35 pm: ah ok thx TT, thought that was a bit weird
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 08:50:06 pm: Quote from: /0 on June 24, 2009, 08:04:35 pm
ah ok thx TT, thought that was a bit weird
Weird about what?
Title: Re: /0's physics phread
Post by: kamil9876 on June 24, 2009, 08:52:45 pm: Quote
[Hard to explain just by typing but imagine the square coil with 2 "faces" the diagram depicts the mag field first hitting the "left face" and then passing through to the "right face", clearly if N to S is positive, the flux would be positive for the "left face" because that what the mag field hits first and as a result the "right face" is negative]

That is correct.

I found that concept far-fetched back in my day but the two faces makes sense if say you pick some rotating reference frame where the coil is stationary and the magnets rotate. That way you know why it is a useful convention to "negativize"(is that even a word?) flux when going through the other face. It's because then the magnets swap(due to rotation) and the flux changes direction and so does current and so the flux to emf relationship holds and you don't have to worry about direction.

Stick to the "two face" concept since it's less confusing in problem solving. But i found the above reasoning helped in adding rigour, being convinced and appreciating the usefulness of the idea.

edit: poor expression and english skills
Title: Re: /0's physics phread
Post by: TrueTears on June 24, 2009, 11:43:46 pm: Quote from: kamil9876 on June 24, 2009, 08:52:45 pm
Quote
[Hard to explain just by typing but imagine the square coil with 2 "faces" the diagram depicts the mag field first hitting the "left face" and then passing through to the "right face", clearly if N to S is positive, the flux would be positive for the "left face" because that what the mag field hits first and as a result the "right face" is negative]

That is correct.

I found that concept far-fetched back in my day but the two faces makes sense if say you pick some rotating reference frame where the coil is stationary and the magnets rotate. That way you know why it is a useful convention to "negativize"(is that even a word?) flux when going through the other face. It's because then the magnets swap(due to rotation) and the flux changes direction and so does current and so the flux to emf relationship holds and you don't have to worry about direction.

Stick to the "two face" concept since it's less confusing in problem solving. But i found the above reasoning helped in adding rigour, being convinced and appreciating the usefulness of the idea.

edit: poor expression and english skills
Asif poor expression, you couldn't have said it any better :)
Title: Re: /0's physics phread
Post by: kamil9876 on June 24, 2009, 11:44:58 pm: after editing :P

I was embarrased by spelling "over" instead of "other". Also I thought the idea may be confusing.
Title: Re: /0's physics phread
Post by: /0 on June 24, 2009, 11:59:52 pm: I understand the two-face concept of the coil, that you can assign 'positive' and 'negative' sides to it. However, say you have a stationary coil with rotating magnets as you described. The stationary coil is lying horizontal and initially the north pole of the magnet is directly above it.
As the magnet rotates around the coil, the sign of the flux when the N pole is above the coil will be different to the sign in flux when the N pole is under the coil. However, without defining the 'sign' of the faces of the coil, we can't decide in which configuration the flux will be positive or negative, right?
Even if we say "let N to S be positive", that doesn't help us because what matters are the 'signs' of the actual face.

... at least, that's what I reckon
Title: Re: /0's physics phread
Post by: TrueTears on June 25, 2009, 12:14:23 am: Quote from: /0 on June 24, 2009, 11:59:52 pm
I understand the two-face concept of the coil, that you can assign 'positive' and 'negative' sides to it. However, say you have a stationary coil with rotating magnets as you described. The stationary coil is lying horizontal and initially the north pole of the magnet is directly above it.
As the magnet rotates around the coil, the sign of the flux when the N pole is above the coil will be different to the sign in flux when the N pole is under the coil. However, without defining the 'sign' of the faces of the coil, we can't decide in which configuration the flux will be positive or negative, right?
Even if we say "let N to S be positive", that doesn't help us because what matters are the 'signs' of the actual face.

... at least, that's what I reckon
It does define what's positive or not.

Think about it, if N to S is positive then the "left face" to "right face" is positive direction and vice versa is negative.
Title: Re: /0's physics phread
Post by: kamil9876 on June 25, 2009, 12:41:50 am: by saying N to S always positive it still makes calculating emf unambigous. u get the same answer (ie: ABCD or DCBA). And i think this is a cool way of thinking about it because it marries what you learnt previously (the fundamentals) without having to introduce some new "face law". (I used it to justify the "face law")

Say it's how u described it, N on top. N to S is positive, hence the flux is down. So once magnet turns 180degrees the flux is up and so the other face is getting hit first, which is the same as the physical situation. We don't have to even look at which face is getting hit first in order to work out emf using this.

Refer back to ussual questions such as, "coil removed from field" and notice the parralels. In fact in such questions you don't worry about faces and it still works.

edit: bahhaa I accidentally talked about keeping magnet stationary :idiot2:
Title: Re: /0's physics phread
Post by: TrueTears on June 25, 2009, 12:43:04 am: Quote from: kamil9876 on June 25, 2009, 12:41:50 am
Say it's how u described it, N on top. N to S is positive, hence the flux is a constant down. So once the coil spins 180degrees the other face is getting hit first, analogous to the conventional, physical situation.
That's right hence it is the opposite sign to what you have defined.
Title: Re: /0's physics phread
Post by: kamil9876 on June 25, 2009, 01:29:07 am: ok so i thought about it a bit:

So N is on top initially and we spin the N magnet anticlockwise(assume constant one radian per second). It can be shown that emf at time $t$ is equal in magnitidue but opposite to direction to emf at time $t+/pi$ . This can be done without any sign conventions, just the plain use of lenze's law(just like coil removed from field problems). However introducing the sign convention that say, flux going down is positive, we see that flux changes sign and using this definition of $\phi(t)$ it can be shown that $\phi'(t+\pi)=-\phi'(t)$ . And so you see it is analogous to what we found out that really happens using just lenze's law without any sign convention. Hence the sign conventions "positive flux is one whose vertial component is down, negative flux is one whose vertical component is up" and "negative emf means in the oppositie direction" coupled with the formula relating flux and emf, yields the correct results that we know are correct from lenze's law.

Interestingly, Once we choose which direction is positive flux and abide the formula $emf=-\phi'(t)$ we automatically choose which direction in the coil is positive emf. If we change our convention of what is negative flux (i.e: $\phi_2=-\phi_1$ ) then we change our direction of what we consider as positive emf. However what is important is the sign changes every $\Delta t=\pi$ . Hence these sign conventions and use of formula are consistent with what we know is really true(what we derived without thinking about sign conventions but just ordinary hand rules)
Title: Re: /0's physics phread
Post by: kamil9876 on June 25, 2009, 01:15:09 pm: ANd yes, i know what you mean now by "N to S is always positive is not enough". In my example what I chose is that Down is ALWAYS positive, as in, "The initial direction of N to S will be always positive". Just like the question in the vcaa exam says: "Take the direction from N to S in the figure as positive". In the figure being analogous to what I mean as "What N to S is initially".
I'm pretty sure that's true as it's the only interpretation that yields only one correct answer as far as our understanding of freedom to assign positive direction goes.
Title: Re: /0's physics phread
Post by: /0 on June 25, 2009, 07:32:31 pm: thanks for all your help guys :D
Title: Re: /0's physics phread
Post by: /0 on June 27, 2009, 05:09:42 am: What is the formula for bright fringes in single-slit diffraction?

And also, the formula for dark fringes is $d\sin{\theta} = n\lambda$

The pattern in single-slit diffraction is a lot b-r-o-a-d-e-r than the pattern in double-slit interference, so how can the same formula apply?
Title: Re: /0's physics phread
Post by: kamil9876 on June 27, 2009, 12:56:40 pm: $d \neq d LOL$ In single slit diffraction, the d ussually means half the width. So when you say broader, what are you keeping the same in both? $d_{double}=w_{single}$ ?

You might like this and the next post: http://vcenotes.com/forum/index.php/topic,9668.msg159096.html#msg159096
The second post explains your first question i hope.
Title: Re: /0's physics phread
Post by: TrueTears on June 27, 2009, 12:57:44 pm: Quote from: /0 on June 27, 2009, 05:09:42 am
What is the formula for bright fringes in single-slit diffraction?

And also, the formula for dark fringes is $d\sin{\theta} = n\lambda$

The pattern in single-slit diffraction is a lot b-r-o-a-d-e-r than the pattern in double-slit interference, so how can the same formula apply?
$sin(\theta) = \frac{\lambda}{w}$ just gives the extent of diffraction. where w is the width of the single slit.
Title: Re: /0's physics phread
Post by: /0 on June 27, 2009, 04:37:06 pm: Quote from: kamil9876 on June 27, 2009, 12:56:40 pm
$d \neq d LOL$ In single slit diffraction, the d ussually means half the width. So when you say broader, what are you keeping the same in both? $d_{double}=w_{single}$ ?

You might like this and the next post: http://vcenotes.com/forum/index.php/topic,9668.msg159096.html#msg159096
The second post explains your first question i hope.

Thanks kamil, I found that post useful
Title: Re: /0's physics phread
Post by: TrueTears on June 27, 2009, 04:40:51 pm: No it gives bright bands for single/double slits.

If you want dark bands it would be $sin(\theta) = \frac{(n-0.5) \lambda}{d}$

[notice it could be n+0.5 due to what you take middle as, ie n = 1 or n = 0]

EDIT: oh you just deleted your last post but anyways I'll leave this up here.
Title: Re: /0's physics phread
Post by: /0 on June 27, 2009, 07:30:56 pm: Thanks TT :)

Also I have another question

$\lambda = \frac{h}{\sqrt{2m_eq_eV}}$ does this take into account relativistic effects?

I thought that electrons usually travel at around $10^6 m/s$
Title: Re: /0's physics phread
Post by: kamil9876 on June 27, 2009, 07:56:44 pm: Quote from: /0 on June 27, 2009, 07:30:56 pm
Thanks TT :)

Also I have another question

$\lambda = \frac{h}{\sqrt{2m_eq_eV}}$ [/tex]

Just out of curiosity, what is that?
Title: Re: /0's physics phread
Post by: /0 on June 27, 2009, 08:05:15 pm: the de broglie wavelength
Title: Re: /0's physics phread
Post by: TrueTears on June 27, 2009, 08:11:28 pm: Quote from: kamil9876 on June 27, 2009, 07:56:44 pm
Quote from: /0 on June 27, 2009, 07:30:56 pm
Thanks TT :)

Also I have another question

$\lambda = \frac{h}{\sqrt{2m_eq_eV}}$ [/tex]

Just out of curiosity, what is that?
$p = \sqrt{2m_eE_k}$
Title: Re: /0's physics phread
Post by: /0 on June 28, 2009, 04:34:07 pm: 1. Why is it that in double-slit experiments we don't consider the contributions from the top and bottom halves of each slit as we do in single-slit diffraction?

2. In single-slit diffraction, you know how they take parallel rays of light and use $\frac{d}{2}\sin{\theta}=\frac{\lambda}{2}$ to find the path difference... well the common reasoning is that the screen is so far away and the slits are so close that the rays are virtually coincident.
But if we make approximations like that, then doesn't that mean that the path difference $\frac{d}{2}\sin{\theta}$ should also be 'insignificant'? I would think that saying two parallel rays converge at infinity is such an impractical estimate that it completely negates the accuracy of $\frac{d}{2}\sin{\theta}$ , which must be VERY accurate if we consider light waves.
Title: Re: /0's physics phread
Post by: TrueTears on June 28, 2009, 04:46:01 pm: 2. Yes, all the formulas are all derived from assuming the point P is at infinity and the 2 lines are parallel, so it is in fact an approximation. But because the width of the slit are so small, the approximation is very very very very close to the path difference.
Title: Re: /0's physics phread
Post by: /0 on June 28, 2009, 06:59:12 pm: Thanks TT, it is a tricky proof

Also, I'm having trouble with VCAA 2003 AOS4 Q5

"In a photoelectric effect experiment, light of various frequencies falls on a metal surface in a photocell. The photoelectrons are decelerated across a retarding voltage, and the stopping potential, $V_s$ , is measured for each frequency. Determine a value for the work function of this metal surface. Include the unit in your answer."

There is a graph which, using regression is approx $V_s = 4.46 \times 10^{-15} f -1.80$ where $V_s$ is in Volts and $f$ in Hz.

The $V_s$ intercept of this graph is $V_s = -1.80V$ .

From $qV_s=hf - W$ , when $f = 0$ , $W = -qV_s$

Therefore, $W = -(1.60 \times 10^{-19})(-1.80) = 2.88 \times 10^{-19} eV$

But if I let $V_s = 0$ then $f=4.04 \times 10^{14}$

And $W = hf = (4.14 \times 10^{-15})(4.04 \times 10^{14}=1.67eV$

Which method is wrong and Why? thanks
Title: Re: /0's physics phread
Post by: kamil9876 on June 28, 2009, 07:05:40 pm: Quote from: /0 on June 28, 2009, 04:34:07 pm
1. Why is it that in double-slit experiments we don't consider the contributions from the top and bottom halves of each slit as we do in single-slit diffraction?

2. In single-slit diffraction, you know how they take parallel rays of light and use $\frac{d}{2}\sin{\theta}=\frac{\lambda}{2}$ to find the path difference... well the common reasoning is that the screen is so far away and the slits are so close that the rays are virtually coincident.
But if we make approximations like that, then doesn't that mean that the path difference $\frac{d}{2}\sin{\theta}$ should also be 'insignificant'? I would think that saying two parallel rays converge at infinity is such an impractical estimate that it completely negates the accuracy of $\frac{d}{2}\sin{\theta}$ , which must be VERY accurate if we consider light waves.

I think you should imagine a single slit not as really just one slit, but a continous domain of infinitely small slits the size of those found in double slit experiments. The reasoning is consistent then. As far as approximations try derive it without approximations, I think you have all the analytical geometry you need at your disposal :P
Title: Re: /0's physics phread
Post by: TrueTears on June 28, 2009, 07:21:40 pm: Quote from: kamil9876 on June 28, 2009, 07:05:40 pm
Quote from: /0 on June 28, 2009, 04:34:07 pm
1. Why is it that in double-slit experiments we don't consider the contributions from the top and bottom halves of each slit as we do in single-slit diffraction?

2. In single-slit diffraction, you know how they take parallel rays of light and use $\frac{d}{2}\sin{\theta}=\frac{\lambda}{2}$ to find the path difference... well the common reasoning is that the screen is so far away and the slits are so close that the rays are virtually coincident.
But if we make approximations like that, then doesn't that mean that the path difference $\frac{d}{2}\sin{\theta}$ should also be 'insignificant'? I would think that saying two parallel rays converge at infinity is such an impractical estimate that it completely negates the accuracy of $\frac{d}{2}\sin{\theta}$ , which must be VERY accurate if we consider light waves.

I think you should imagine a single slit not as really just one slit, but a continous domain of infinitely small slits the size of those found in double slit experiments. The reasoning is consistent then. As far as approximations try derive it without approximations, I think you have all the analytical geometry you need at your disposal :P
True, I imagine a single slit having 2 "corners" which the wave has to bend around hence cause diffraction. I think that is a good way of thinking about it.
Title: Re: /0's physics phread
Post by: kamil9876 on June 28, 2009, 07:28:37 pm: Whoa, looked like a mathematical inconsistency haha.

$E=hf-W$
$V_s=\frac{h}{q_e}f-\frac{W}{q_e}$

Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of $q_e$ only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be $\frac{h}{q_e}$ this is not actually the case according equal to the values of $q_e$ and $h$ that you have assumed.
Title: Re: /0's physics phread
Post by: TrueTears on June 28, 2009, 07:35:57 pm: Quote from: /0 on June 28, 2009, 06:59:12 pm
Thanks TT, it is a tricky proof

Also, I'm having trouble with VCAA 2003 AOS4 Q5

"In a photoelectric effect experiment, light of various frequencies falls on a metal surface in a photocell. The photoelectrons are decelerated across a retarding voltage, and the stopping potential, $V_s$ , is measured for each frequency. Determine a value for the work function of this metal surface. Include the unit in your answer."

There is a graph which, using regression is approx $V_s = 4.46 \times 10^{-15} f -1.80$ where $V_s$ is in Volts and $f$ in Hz.

The $V_s$ intercept of this graph is $V_s = -1.80V$ .

From $qV_s=hf - W$ , when $f = 0$ , $W = -qV_s$

Therefore, $W = -(1.60 \times 10^{-19})(-1.80) = 2.88 \times 10^{-19} eV$

But if I let $V_s = 0$ then $f=4.04 \times 10^{14}$

And $W = hf = (4.14 \times 10^{-15})(4.04 \times 10^{14}=1.67eV$

Which method is wrong and Why? thanks

Your first method is wrong.

it is $q_eV_s = -W$

$1 \times -1.80 = -W$

$W = 1.8eV$

Your 2nd method is fine.

The reason 1.8eV does not match with your ~1.7eV answer is because of the estimated intercepts, both answers are fine.
Title: Re: /0's physics phread
Post by: TrueTears on June 28, 2009, 08:06:38 pm: Quote from: kamil9876 on June 28, 2009, 07:28:37 pm
Whoa, looked like a mathematical inconsistency haha.

$E=hf-W$
$V_s=\frac{h}{q_e}f-\frac{W}{q_e}$

Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of $q_e$ only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be $\frac{h}{q_e}$ this is not actually the case according equal to the values of $q_e$ and $h$ that you have assumed.
/0's inconsistency in his/her answers was not due to the use of gradient or whatever but it was because he/she did not recognize that he/she got mixed up with eV and Joules.

ok.
Title: Re: /0's physics phread
Post by: kamil9876 on June 28, 2009, 09:18:34 pm: Quote from: TrueTears on June 28, 2009, 08:06:38 pm
Quote from: kamil9876 on June 28, 2009, 07:28:37 pm
Whoa, looked like a mathematical inconsistency haha.

$E=hf-W$
$V_s=\frac{h}{q_e}f-\frac{W}{q_e}$

Equating the constant term in the end with what they give you gives you're first method. Notice that this requires knowledge of $q_e$ only, and does not require knowledge of the gradient they have given you.

Your second method does use the gradient they have given you hence any errors that have crept in must have been caused by this. Notice that that gradient must be $\frac{h}{q_e}$ this is not actually the case according equal to the values of $q_e$ and $h$ that you have assumed.
/0's inconsistency in his/her answers was not due to the use of gradient or whatever but it was because he/she did not recognize that he/she got mixed up with eV and Joules.

ok.

Quote
Notice that that gradient must be \frac{h}{q_e} this is not actually the case according equal to the values of q_e and h that you have assumed.

This pretty much says that $q_e$ or $h$ is wrong. Because even if he/she/it derived the gradient using WRONG values of q_e or h, he/she/it would find no inconsistency in the two methods. But the occurence of an inconsistency suggests a mistake in either the given gradient or the two constants. Hence I pointed him/her/it to the sauce of the error.
Title: Re: /0's physics phread
Post by: /0 on June 28, 2009, 09:45:14 pm: lol thanks guys, gah I had the wrong units for q
Title: Re: /0's physics phread
Post by: TrueTears on June 28, 2009, 09:47:14 pm: Quote from: /0 on June 28, 2009, 09:45:14 pm
lol thanks guys, gah I had the wrong units for q
took you a while to figure that out.
Title: Re: /0's physics phread
Post by: /0 on June 28, 2009, 11:57:58 pm: Quote from: TrueTears on June 28, 2009, 09:47:14 pm
Quote from: /0 on June 28, 2009, 09:45:14 pm
lol thanks guys, gah I had the wrong units for q
took you a while to figure that out.

yes i am quite stupid

Quote from: TrueTears on June 28, 2009, 08:06:38 pm
/0's inconsistency in his/her answers

???
Title: Re: /0's physics phread
Post by: /0 on June 29, 2009, 02:13:48 am: In Nelson there's

$\sin{\theta} = \frac{x}{L}$

$\sin{\theta} = \frac{\lambda}{d}$

$\Rightarrow \frac{x}{L} = \frac{\lambda}{d}$

Where 'L' is the distance from the slits to the screen.

However, in the derivation the set up attached is used. This does not have 'L' as the perpendicular distance.

Did the book assume $\theta$ to be small, so that $L \approx perpendicular \ distance$ ?

However, isn't this made very inaccurate by the assumption that the screen is very far away?
Title: Re: /0's physics phread
Post by: TrueTears on June 29, 2009, 02:32:12 am: Yes, I had the exact same problem, I think book assumes theta is very small hence L is the vertical distance :P
Title: Re: /0's physics phread
Post by: /0 on June 29, 2009, 03:29:22 am: Cool
also http://en.wikipedia.org/wiki/Planck_length

Apparently that's the smallest length possible. But what if you have rock of $1kg$ moving at $10^3m/s$ , then

$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(1)(10^3)} =$ .

Moderator action: post edited by Mao
Title: Re: /0's physics phread
Post by: TrueTears on June 30, 2009, 12:34:15 am: Eh what's all that spam for???

All I see is boxes.

Stop spamming. Your trolling needs to stop it is getting annoying.
Title: Re: /0's physics phread
Post by: /0 on June 30, 2009, 05:43:35 am: nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?
Title: Re: /0's physics phread
Post by: NE2000 on June 30, 2009, 09:28:07 am: Quote from: /0 on June 30, 2009, 05:43:35 am
nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?

Sound is reflected out of phase at open ends of a pipe and in the same phase at closed ends of a pipe. That is, at an open end a compression will turn into a rarefaction and vice versa. Therefore at the open end there will be destructive interference occurring. Hence a node.
Title: Re: /0's physics phread
Post by: /0 on June 30, 2009, 01:03:25 pm: Quote from: NE2000 on June 30, 2009, 09:28:07 am
Quote from: /0 on June 30, 2009, 05:43:35 am
nevermind, I thought you would get it (also afaik i haven't trolled in a while)

A very large pipe in a science museum is open at both ends. It is large enough for people to walk inside it. A large loudspeaker faces inwards on the right end. Jin is walking along the pipe.

Later it goes on to say "the two ends of the pipe are nodes".

If the pipe is open at both ends then shouldn't there be antinodes at both ends?

Sound is reflected out of phase at open ends of a pipe and in the same phase at closed ends of a pipe. That is, at an open end a compression will turn into a rarefaction and vice versa. Therefore at the open end there will be destructive interference occurring. Hence a node.

thanks NE2000 :)
Title: Re: /0's physics phread
Post by: /0 on August 06, 2009, 08:38:44 pm: When sketching field lines around a bar magnet do you have to sketch the field inside the magnet?

How do you convert 5920kWh into GJ? I keep getting the wrong answer.

A magnet falls through the centre of a coil. At one stage the current through the $10\Omega$ resistor is $0.020A$ . Which of the following is closest to the rate of change of the magnetic field at this time?
a) $0.001 Wbs^{-1]$
b) $0.002Wbs^{-1}$
c) $0.004Wbs^{-1}$
d) $0.2Wbs^{-1}$

thanks!
Title: Re: /0's physics phread
Post by: TrueTears on August 06, 2009, 08:42:27 pm: for 2nd question didn't you do this at school today?

(I think you forgot the number of turns man, check the first part :P 50 turns!!!)

for first question... $1khw = 3.6 \times 10^6 J$ , you can work it out from there.
Title: Re: /0's physics phread
Post by: /0 on August 07, 2009, 07:39:41 am: lol thanks tt